Pressure – Volume – Temperature

Relationship of Pure Fluids

Volumetric data of substances are needed to calculate the thermodynamic

properties such as internal energy and enthalpy, from which the heat and

work requirements of processes are obtained

Understanding the P-V-T behavior of pure substances allows the engineer to

make accurate estimations to the changes in properties accompanying their

state changes

Pressure, volume and temperature are experimentally controllable

properties that are connected by equation of state

A pure substance can have as many as 3 phases coexisting at the same

conditions: gas, liquid, solid

P-V-T behavior of any substance can be experimentally obtained and

represented in the form of phase diagrams

𝑉 𝑃, 𝑇

𝑉 = 𝑅𝑇/𝑃 For ideal gas

The word fluid covers both gases and liquids, any gas and any liquid is a

fluid

A phase is considered a liquid if it can be vaporized by reduction in pressure

at constant temperature

A phase is considered a gas if it can be condensed by reduction of

temperature at constant pressure

Vapor is the preferred term for gas when the gaseous phase is in equilibrium

with the corresponding liquid or solid, although there is no significant

physical or chemical difference between a vapor and a gas

Vaporization curve in a P-T diagram of a pure substance seperates vapor and

liquid phases upto a critical temperature and pressure where the curve ends

Above the critical temperature and/or critical pressure, fluid is the

preferred term, because there is no meaningful distinction between

liquid/vapor/gas

Pressure – Temperature diagram of water

It is convinient to call the phase below critical pressure the gas phase and the gas

phase below critical temperature the vapor phase which can be condensed to liquid

vapor

Pressure – Volume diagram of a pure fluid Isotherms greater than the critical temperature do not cause phase changes and

therefore are smooth

Isotherms lower than the critical temperature consist of three distinct regions:

vapor, liquid-vapor, liquid

• At the horizontal sections which represent

the phase change between vapor and liquid,

the constant pressure is the vapor pressure

• Ratio of liquid and vapor phases can be

obtained by the lever rule

• The curve B-C represents saturated liquid

• The curve C-D represents saturated vapor

• Isotherms in the liquid region are steep

because liquid volumes change little with

large changes in pressure

• The horizontal sections of the isotherms

get gradually shorter with increasing T,

approaching point C

• The liquid and vapor phases have the same

properties at the horizontal inflection point at C

Temperature - Volume diagram of a pure fluid

Temperature increase is proportional to volume except the phase change

region

A relation connecting pressure, volume and temperature exists for the

regions of the diagrams where single phase exists:

𝑓 𝑃, 𝑉, 𝑇 = 0

or

The partial derivatives in this equation have definite physical meanings and

measurable quantities for fluids

The volume expansion coefficient

The isothermal compression coefficient

Combining provides the general equation relating P, V and T for fluids:

d𝑉 = 𝜕𝑉

𝜕𝑇 𝑃𝑑𝑇 +

𝜕𝑉

𝜕𝑃 𝑇𝑑𝑃

𝛽 = 1

𝑉

𝜕𝑉

𝜕𝑇 𝑃

𝜅 = −1

𝑉

𝜕𝑉

𝜕𝑃 𝑇

d𝑉

𝑉= 𝛽𝑑𝑇 − 𝜅𝑑𝑃

The isotherms for the liquid phase on the left side of the P-V and T-V

diagrams are steep and closely spaced which means both volume expansion

and isothermal compression coefficients of liquids are small

A useful idealization known as incompressible fluid is employed in fluid

mechanics for a sufficiently realistic model of liquid behavior

The volume expansion and isothermal compression coefficient of the

incompressible fluid are zero so it cannot be described by an equation of

state relating V to T and P

β and κ are weak functions of temperature and pressure for real liquids

Thus error introduced by taking them constant is low for small changes in T

and P

ln𝑉2

𝑉1= 𝛽(𝑇2 − 𝑇1) − 𝜅(𝑃2 − 𝑃1)

d𝑉 = V𝛽𝑑𝑇 − 𝑉𝜅𝑑𝑃

Example – The volume expansion and isothermal compression coefficients of

water at room temperature and 1 atm pressure are given as:

β= 0.257 x 10-3 °C-1

κ= 4.6 x 10-5 atm-1

and V = 1.00296 cm³ for 1 gram

• Calculate the pressure generated when water is heated at constant

volume from 25°C and 1 atm to 40°C

V, β and κ are assumed constant in the small temperature interval

𝑃2 = 𝑃1 + ∆𝑃 = 1 + 83.8 = 84.8 atm

ln𝑉2

𝑉1= 0 = 𝛽(𝑇2 − 𝑇1) − 𝜅(𝑃2 − 𝑃1)

∆𝑃 =𝛽

𝜅∆𝑇 = 5.587 ∗ 15 = 83.8 atm

• Obtain the value of 𝜕𝑃

𝜕𝑇 𝑉

• Calculate the volume change when the state of water is changed from

25°C and 1 atm to 0°C and 10 atm

β and κ are assumed constant in the small temperature and pressure

interval

𝑉2𝑉1= 0.9932

𝑉2 = 0.9961 cm³ for 1 gram

∆𝑉 = −0.00682 cm³ for 1 gram

ln𝑉2𝑉1= 0.257 ∗ 10−3 −25 − 4.6 ∗ 10−5 9

= −0.006839

𝜕𝑃

𝜕𝑇𝑉

=𝛽

𝜅=0.257 ∗ 10−3

4.6 ∗ 10−5= 5.587

atm

°C

𝑑𝑉

𝑉= 0 = 𝛽𝑑𝑇 − 𝜅𝑑𝑃

𝑃𝑉 = 𝑎 + 𝑏𝑃 + 𝑐𝑃2 +⋯

𝑃𝑉 = 𝑎(1 + 𝐵𝑃 + 𝐶𝑃2 +⋯)

The P-V-T relationship of fluids is hard to formulate

However relatively simple equations can describe

the P-V-T behavior of gas regions

For gases P decreases proportionally to

the increase in V

Thus 𝑃 ∝ 1𝑉 , 𝑃𝑉 ≈ 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

PV along an isotherm is represented by a power

series expansion

where a, B, C, etc are constants for a given

temperature and chemical species

Parameter a is a function of temperature for all species

The relationship is truncated for simplicity

In general, the greater the pressure range the larger the number of terms required

Data taken for various gases at a

specific constant temperature show

that P-V plots have the same limiting

value of PV/T as P0

a is the same for all gases!

This remarkable property of gases

helped scientists establish a

temperature scale that is

independent of the gas used as

thermometric fluid:

lim𝑃=0

𝑃𝑉 ≡ (𝑃𝑉)𝑜 = 𝑎 = 𝑓(𝑇)

𝑃𝑉 𝑜 = 𝑎 = 𝑅𝑇

𝑃𝑉 𝑜𝑡= 𝑅 ∗ 273.16 K where the t denotes the value of the triple point of water

𝑇 K = 273.16 𝑃𝑉 𝑜

𝑃𝑉 𝑜𝑡

Thus Kelvin or ideal gas temperature scale established throughout the temperature

range for which limiting values of PV as P0 are experimentally accessible

where R denotes the universal gas constant

T=273.16 K

𝑃𝑉 = 𝑎(1 + 𝐵𝑃 + 𝐶𝑃2 +⋯)

Constant a is replaced by RT with the establishment of ideal gas

temperature scale

where the ratio PV/RT, Z is called the compressibility factor

An alternative representation of Z:

B, C, D, etc called virial coefficients represent the interaction between

pairs of molecules, three-body interactions, four-body interactions

respectively

Contributions to Z of the higher-ordered terms fall off rapidly

As the pressure of a real gas is reduced at constant temperature, its volume

increases and the contributions of the terms B’/V, C’/V2 decrease

As pressure of the gas approaches zero, molecular distance become infinite,

molecular attractions become negligible and Z approaches unity

so

𝑃𝑉 = 𝑎(1 + 𝐵𝑃 + 𝐶𝑃2 + 𝐷𝑃3 +⋯)

𝑍 =𝑃𝑉

𝑅𝑇= 1 + 𝐵𝑃 + 𝐶𝑃2 + 𝐷𝑃3 +⋯

𝑍 = 1 +𝐵′

𝑉+𝐶′

𝑉2+𝐷′

𝑉3+⋯

𝑃𝑉 = 𝑅𝑇 for 1 mole of ideal gas

The use of the compressibility factor and the associated equation expansion

is practical when no more than two or three terms are required

This requirement is satisfied for gases at low to moderate pressures

The plots of compressibility factors of gases as a function of pressure at

various temperatures show that Z=1 at P=0 and the isotherms are nearly

straight at low to moderate pressures

The tangent to an isotherm at P=0 is a good approximation of the isotherm

for a finite pressure range

or

𝑍 = 1 + 𝐵𝑃 + 𝐶𝑃2 + 𝐷𝑃3 +⋯

𝑑𝑍

𝑑𝑃= 𝐵 + 2𝐶𝑃 + 3𝐷𝑃2 +⋯

𝑑𝑍

𝑑𝑃𝑃=0

= 𝐵

𝑍 = 1 + 𝐵𝑃

𝑍 = 1 +𝐵′

𝑉

Represent the PVT behavior of most gases up to a P of 15bar

Applicable to gases below pressures of 50 bar

The equation is cubic in volume so solution for V is usually done by an

iterative way

Example – Calculate the volume and compressibility factor of isopropanol

vapor at 200 °C and 10 bar. The virial coefficients of isopropanol at 200 °C

is given as

B’= -388 cm3 mol-1, C’= -26000 cm6 mol-2

Calculate by

• The ideal gas equation

• 𝑍 = 1 +𝐵′𝑃

𝑅𝑇

• Z = 1 +𝐵

𝑉+

𝐶

𝑉2

𝑍 =𝑃𝑉

𝑅𝑇= 1 +

𝐵′

𝑉+𝐶′

𝑉2

𝑍 = 1 +𝐵

𝑉

𝐵 = 𝐵′ 𝑅𝑇

𝑍 = 1 +𝐵′𝑃

𝑅𝑇

The compressibility factor accurately describes the PVT behavior of gases at

low to medium pressures

A cubic equation of higher complexity is applied to define the PVT behavior

of liquids as well as gases

The equation also known as Redlich/Kwong equation has three roots at most

of the low pressure range

The smallest root is liquid volume, the highest vapor volume and the middle

is of no significance

Thus volumes of saturated liquid and saturated vapor are given by the

smallest and largest roots when P is the saturation or vapor pressure

The equation is solved by iteration

See text book for iteration steps and calculation of the constants a and b

𝑃 =𝑅𝑇

𝑉 − 𝑏−

𝑎

𝑇12 𝑉(𝑉 + 𝑏)

Ideal gas equation of state is a model equation applicable to all gases to

understand their P-V-T behavior and the energy requirements of processes

within small margins of error

Consider an engine piston full of ideal gas

Total energy of the ideal gas can only be changed through transfer of energy

across its boundary and work done on or by the system

Heat is admitted to the system by conduction through metal block

Work is done on the system by compressing the piston

Work is done by the system by expansion of the piston

Let U be a thermodynamic state function of the system called the internal energy

ΔU is the change in internal energy of the system for any process

Q is the quantity of heat that flows into the system during process

W is defined as the mechanical work done on the system by the external pressure

W’ is defined as all other kinds of work done on the system during the process

Thus

U is a state function and is independent of the path

Q,W, and W’ are process variables that depend on the path

Mechanical work is the only work done in practical applications and W’ is omitted

The signs of Q and W depend on the direction of the energy transfer

They are positive when heat is transferred into the system and work is done by it

through expansion

They are negative when heat flows out of the system and work is done on it by

compression

∆𝑈 = 𝑄 +𝑊 +𝑊′

∆𝑈 = 𝑄 −𝑊

Mathematical formulation of the

First law of thermodynamics

Specific Heat:

It is the amount of heat required to raise the temperature of a 1 kg mass

1C or 1K

KJ )T - (T mC Q

m, mass thegConsiderin

kg

KJ )T - (T C Q

nintegratioby

CdT dQ

12

12

dt

dQC

Example – A batch of 20 tons of liquid copper is to be cooled from 1200 °C

to 1150 °C by adding solid copper at 25 C. 40000 kJ are lost during the time

it takes to make the addition and the temperature to stabilize. What

quantity of solid is used?

The Constant Volume Process

Internal energy of the ideal gas only changes by heat flow when the piston

is not allowed to expand or compress

where CV is the molar constant-volume heat capacity of the gas

ΔU and CV of ideal gases are functions of only temperature so

ΔU for an ideal gas can always be calculated by 𝐶𝑉𝑑𝑇 , regardless of the

process

𝑑𝑈 = 𝑑𝑄 = 𝐶𝑉𝑑𝑇

∆𝑈 = 𝑄 −𝑊

The Constant Pressure Process

The first law of thermodynamics simplifies accordingly for the special case of

constant P:

∆𝑈 = 𝑄 −𝑊

∆𝑈 = 𝑄 − 𝑃∆𝑉

𝑄 = ∆𝑈 + 𝑃∆𝑉 state function

Enthalpy change ∆𝐻 = 𝑄 for a constant P process

𝐶 =𝛿𝑄

𝑑𝑇

𝑑𝐻 = 𝛿𝑄 = 𝐶𝑝𝑑𝑇

𝐻 = 𝑈 + 𝑃𝑉 = 𝑈 + 𝑅𝑇

Both enthalpy and 𝐶𝑝 of ideal gases also depend only on temperature

𝑑𝐻 = 𝐶𝑝𝑑𝑇 for all process of ideal gas just as 𝑑𝑈 = 𝐶𝑉𝑑𝑇

𝑑𝐻 = 𝑑𝑈 + 𝑃𝑑𝑉

𝑑𝐻 = 𝑑𝑈 + 𝑅𝑑𝑇

𝐶𝑝𝑑𝑇 = 𝐶𝑉𝑑𝑇 + 𝑅𝑑𝑇

𝐶𝑝 − 𝐶𝑉 = 𝑅 for ideal gas

The Constant Temperature Process

Internal energy of ideal gas stays constant in an isothermal process

𝑑𝑈 = 𝑑𝑄 − 𝑑𝑊 = 0

𝑄 = 𝑊

For a mechanically reversible process work is done by expansion

𝑊 = 𝑃𝑑𝑉 = 𝑅𝑇𝑑𝑉

𝑉

𝑄 = 𝑊 = 𝑅𝑇 ln𝑉2𝑉1

Since 𝑃1 𝑃2 = 𝑉2 𝑉1 for constant temperature,

𝑄 = 𝑊 = 𝑅𝑇 ln𝑃1𝑃2

T

The Adiabatic Process

There is no heat flow into or out of the system in adiabatic processes

𝑑𝑈 = −𝑑𝑊 = 𝑃𝑑𝑉

Since internal energy change of ideal gas for any process is 𝑑𝑈 = 𝐶𝑉𝑑𝑇,

𝐶𝑉𝑑𝑇 = −𝑃𝑑𝑉 = −𝑅𝑇

𝑉𝑑𝑉

𝑑𝑇

𝑇=−𝑅

𝐶𝑉

𝑑𝑉

𝑉

integrating,

𝑅𝐶𝑣 =

𝐶𝑝

𝐶𝑣− 1 since 𝐶𝑝 − 𝐶𝑣 = 𝑅 for ideal gases

𝐶𝑣ln𝑇2𝑇1

= 𝑅ln𝑉1𝑉2

𝑇2𝑇1

=𝑉1𝑉2

𝑅𝐶𝑣

𝑇2𝑇1

=𝑃2𝑉2𝑃1𝑉1

=𝑉1𝑉2

𝐶𝑝𝐶𝑣−1

𝑃2𝑃1

=𝑉1𝑉2

𝐶𝑝𝐶𝑣

𝑃1𝑉1

𝐶𝑝𝐶𝑣 = 𝑃2𝑉2

𝐶𝑝𝐶𝑣 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

Remember the reversible isothermal expansion of an ideal gas

𝑄 = 𝑊 = 𝑃𝑑𝑉 = 𝑅𝑇ln𝑉2𝑉1

= 𝑅𝑇ln𝑃1𝑃2

so 𝑃1𝑉1 = 𝑃2𝑉2 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

The work done by an reversible isothermal process exceeds that of the reversible

adiabatic because the internal energy of the adiabatically contained system

decreases while performing work

Isothermal paths are utilized in heat engines to perform higher work

The General Process

When no specific conditions other than mechanical reversibility are present, the

following general equations apply to ideal gas:

𝑑𝑈 = 𝑑𝑄 − 𝑑𝑊 ∆𝑈 = ∆𝑄 − ∆𝑊

𝑑𝑊 = 𝑃𝑑𝑉 ∆𝑊 = 𝑃𝑑𝑉

𝑑𝑈 = 𝐶𝑉𝑑𝑇 ∆𝑈 = 𝐶𝑉𝑑𝑇

𝑑𝐻 = 𝐶𝑃𝑑𝑇 ∆𝐻 = 𝐶𝑃𝑑𝑇

Values for Q cannot be determined directly and is obtained from the first law:

𝑑𝑄 = 𝐶𝑉𝑑𝑇 + 𝑃𝑑𝑉

𝑄 = 𝐶𝑉𝑑𝑇 + 𝑃𝑑𝑉

Example – A closed system undergoes a process, during which 1000 J of heat is

removed from it and 2500 J of work is done by it. The system is then restored to its

initial state. If1500 J of heat addition is required for the second process, what is the

amount of work done by or to the system in the second process?

∆𝑈 = ∆𝑄 − ∆𝑊

Example – Most real gases obey the following equation of state:

𝑃𝑉 = 𝑛𝑅𝑇 + 𝑏𝑃 Find an expression for work when the gas undergoes a reversible isothermal volume

expansion

𝑑𝑊 = 𝑃𝑑𝑉 ∆𝑊 = 𝑃𝑑𝑉

Example – The general relationship between P-V-T of a solid is

V = 𝑉0 1 + 𝛽∆𝑇 − 𝜅∆𝑃 where subscript 0 represents a reference state. 𝛽 for copper is 5*10-5 K-1 and 𝜅 is 7*10-7 bar-1. For reference state P0=1 bar, T0 is 300 K and the density of copper is

0.94 g/cm3. 40 kJ of heat is added to a 5 kg block of copper starting at this state, at

constant pressure. The resulting increase in temperature is 200 C. Determine the

amount of work done and the change in internal energy.

Example – What pressure increase is needed to make a block of magnesium metal

retain its initial volume while it is being heated from 0 to 50 °C? The metal block is

cubic in shape

𝛽 for magnesium is 2.5*10-7 K-1 and 𝜅 is 2.95*10-1 bar-1.

V = 𝑉0 1 + 𝛽∆𝑇 − 𝜅∆𝑃